int {{x^x}(1 + ln x)dx} is equal to :- | vedclass

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(A) Let $I = \int x^x (1 + \ln x) dx$.Consider the function $f(x) = x^x$.To differentiate $f(x)$,take the natural logarithm on both sides:$\ln f(x) =

Vedclass · Accepted Answer

$x^x + C$

Vedclass · Answer

(A) Let $I = \int x^x (1 + \ln x) dx$.Consider the function $f(x) = x^x$.To differentiate $f(x)$,take the natural logarithm on both sides:$\ln f(x) = \ln(x^x) = x \ln x$.Differentiating both sides with respect to $x$:$\frac{1}{f(x)} f'(x) = \frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.Therefore,$f'(x) = f(x) (1 + \ln x) = x^x (1 + \ln x)$.Since the integrand is of the form $\int f'(x) dx$,the integral is $f(x) + C$.Thus,$I = x^x + C$.

$\int {{x^x}(1 + \ln x)dx} $ is equal to :-

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